the fourier series expansion of a real periodic signal with fundamental frequency f0 is given by

26. The Laplace transform of i(t) is given by i(s) = 2/s(1 + s). as t the value of i(t) tends to

(a) 0

(b) 1

(d) 0

The Fourier series expansion of a real periodic signal with fundamental frequency f_o is given by g_p(t) = c₀e^j2nft . it is given that c₃ = 3 + j5.

(a) 5 + j3

(b) -3 – j5

(d) 3 – j5

Let x(t) be the input to a linear, time-invariant system. the required output is 4x(t – 2). the transfer function of the system should be

(a) 4e^j4t

(b) 2e^-j8t

(d) 2e^j8t

A sequence x(n) with the z-transform x(z) = z⁴ + z² – 2z + 2 – 3z^-4 is applied as an input to a linear, time-invariant sysyem with the impulse response h(n) = 2(n – 3), where (n) = {1, n = 0 0, otherwise

the output at n = 4 is

(a) -6

(b) zero

(d) -4

Convolution of x(t + 5) with impulse function (t – 7) is equal to

(a) x(t – 12)

(b) x(t + 12)

(d) x(t + 2)

Which of the following cannot be the fourier series expansion of a periodic signal?

(a) x(t) = 2 cos t + 3 cos 3t

(b) x(t) = 2 cos t + 7 cos t

(d) x(t) = 2 cos 1.5t + sin 3.5 t

The fourier transform f(e^-t u(t) is equal to 1/1 + j2f. therefore, f{1/1 – j2t} is

(a) e^f u(f)

(b) e^-f u(f)

(d) e^-f u(-f)

A linear phase channel with phase delay t_p and group delay t_g must have

(a) t_p = t_g = constant

(b) t_p f and t_g f

(d) t_p f and t_g = constant

Conider a sampled signal :

y(t) = 5 x 10^-6 x(t) (t – nt_s),

where, x(t) = 10 cos (8 x 10³) t

and t_s = 100 us. when y(t) is passed through an ideal low-pass filter with a cut-off frequency of 5 khz, the output of the filter is

(a) 5 x 10^-6 cos (8 x 10³) t

(b) 5 x 10^-5 cos (8 x 10³) t

(d) 10 cos (8 x 10³) t

The transfer function of a system is given by h(s) = 1/s² (s – 2) . the impulse response of the system is (* denotes convolution, and u(t) is unit step function)

(a) (t² * e^-2t) u(t)

(b) (t * e^2t) u(t)

(d) (te^-2t) u(t)

The region of convergence of the z-transform of a unit step function is

(a) |z|>1

(b) |z|<1

(d) (real part of z) < 0

Let (t) denotes the delta function. the value of the integral (t) cos (3t/2) dt is

(a) 1

(b) -1

(d) 2

A band limited signal is sampled at the nyquist rate. the signal can be recovered by passing the samples through

(a) an R-C filter

(b) an envelope detector

(d) an ideal low-pass filter with appropriate bandwidth

A linear time-invariant system has an impulse response e^2t, t > 0. if the initial conditions are zero and the input is e^3t, the output for t > 0 is

(a) e^3t – e^2t

(b) e^5t

(d) none of these

Given that l[f(t)] = s + 2/s² + 1, l[g(t)] = s² + 1/(s + 3) (s + 2). h(t) = f(r) g(t) d. l[h(t)] is

(a) s²+ 1/s + 3

(b) 1/s + 3

(d) none of these

The fourier transform of the signal x(t) = e^-3t2 is of the following form, where A and B are constants

(a) Ae^-5(t)

(b) Ae^-bf2

(d) Ae^-bf

A system with an input x(t) and output y(t) is described by the relation y(t) = tx(t). this system is

(a) linear and time-invariant

(b) linear and time-variant

(d) non-linear and time-variant

The n-point discrete fourier transform of a signal x[n] is x_dft [k]. if DFT of another discrete sequence y[n] is given as y_dft [k] = x*_dft [k], then which of the following is true?

(a) y[n] = x* [n]

(b) y[n] = nx* [n]

(d) y[n] = – x* [n]

Discrete fourier transform of a real sequence {x[n]} is given as

x_dft[k] = {0, a, 2 + j, – 1,b,j}

then, a and b are

(a) 2 – j, j

(b) j, -j

(d) j, 2 + j

The 4-point discrete fourier transform of a discrete time sequence (1,2,1,2) is

(a) {0, -2, 0, 6}

(b) {6, -4j, 6 – 4j}

(d) {6, 0, -2, 0}

Consider a band-pass signal x_c (t) is shown below. the minimum sampling rate required for this signal to prevent loss of information is

(a) 4 kHz

(b) 12 kHz

(d) 2 kHz

Two signals x₁(t) are band-limited to 2 kHz and 3kHz respectively, then nyquist rate for the signal x₁(t)* x₂(t) is

(a) 5 kHz

(b) 6 kHz

(d) 10 kHz

Which one is most appropriate dynamic system out of the following?

(a) y(n) = y(n – 1) + y(n + 1)

(b) y(n) = y(n – 1)

(d) y(n) + y(n – 1) + y(n + 3) = 0

Which one is causal system?

(a) y(n) = 3x(n) – 2x(n – 1)

(b) y(n) = 3x(n) + 2x(n + 1)

(d) y(n) = 3x(n + 1) + 2x(n – 1) + x (n)

For the signal given below

y(t) = 2 sin (2/3 t) + 4sin (1/4 t 4) + 6 sin (1/3 t 5) + 8 sin (1/2 t 7)

the common period of y(t) is given by

(a) 12

(b) 24

(d) 16

from final value theorem,

lim I(t) = lim sI (s)

= lim s x 2/s (1 + s)

= 2

the coefficients c_n and c_-n in the fourier series are complex conjugate.

C_-N = C_N*

|C_-N |= |C_N|

<C_-N = – <C_N

C₃ = 3 + J5

C_-3 = C₃* = 3 – J5

transfer function h(s) = FT of output/FT if input

= 4e^-j20 x(s)/x(s) = 4e^-j2o

h(s) = 4e^-j4f

y(z) = h(z) x(z)

given, x(z) = z⁴ + z² – 2z + 2 – 3z^-4

h(n) = [n – 3]

h(z) = z^-3

y(z) = z^-3 [z⁴ + z² – 2z + 2 – 3z^-4]

= z + z^-1 + 2z^-2 + 2z^-3 – 3z^-7

taking inverse z-transform,

by inspection,

y[n] = [n + 1] + [n – 1] – 2[n – 2] + 2 [n – 3] – 3 [n – 7]

putting n = 4

y[4] = 0

from shifting property of impulse function,

(t)* (t – t_o) = (t – t₀)

hence, x(t + 5)* (t – 7) = x(t + 5 – 7)

= x(t – 2)

signal x(t) = 2 cos t + 7 cos t

period of 2 cos t = t₁ = 2/0

= 2 = 2

period of 7 cos t = t₂ = 2/0

= 2/1 = 2

for signal x(t) to be periodic t₁/t₂ not be rational.

t₁/t₂ = 2/2 = 1 = irrational

periodic waveform is sufficient condition for existence of fourier series. hence, x(t) = 2 cos t + 7 cos t is not a periodic signal hence, fourier series expansion is not possible.

from duality property of fourier transform.

x(t) – x(f)

x(t) – x(-f)

f{1/1 + j2t} = e^f u(-f)

let V_I = x(t) cos t is applied in linear phase channel.

where, T_g = group delay

T_p = phase delay

in liear phase channel, transmission is distrtionless.

hence, T_P = T_G = constant

given that sampling interval t_s = 100 us

sampling frequency f_s = 1/t_s = 10 kHz

bandwidth of LPF = f_c = 5 kHz

f_s = 2 B

hence, sampled output not overlap.

filter output = y(t)/t_s

= 5 x 10^-6 x x(t)/100 x 10^-6

given, x(t) = 10 cos (8 x 10³) t

output = 5 x 10^-6 x 10 cos (8 x 10³) t/100 x 10^-6

= 5 x 10^-1 cos (8 x 10³) t

transfer function h(s) = 1/s²(s – 2)

lmpulse response h(t) = l^-1 [h(s)]

= l^-1 [1/s²(s – 2)]

we know that

l^-1 [f₁ (t). f₂(t)] = l^-1 [f₁(t)* l^-1[f₂(t)]

hence, h(t) = l^-1 [1/s²]* l^-1|1/s – 2]

= t u(t)* e^2t u(t)

= (t^* e^2t) u(t)

x[n] = u[n]

x[z] = x[k] z^-k

u[k] z^-k = z^-k

x[z] = 1/1 – z^-1

for x[z] to converge,

|z^-1|< 1

|z|> 1

from sampling property of unit impulse function,

(t) (t) dt = (0)

(t) cos (3t/2) dt = cos (0) = 1

As band limited signal is sampled at nyquist rate. hence, replicated spectra do not overlap and the original sepctrum can be generated by passing 0_s (t) through ideal low-pass filter that has cut-off frequency.

(bandwidth f_c = f_s/2 ; where f_s is sampling frequency.

y(s) = x(s) h(s)

given, x(t) = e^3t, h(t) = e^2t

x(s) = 1/s – 3, h(s) = 1/s – 2

y(s) = x(s) h(s) = 1/(s – 3) x 1/(s – 2)

= 1/s – 3 – 1/s – 2

hence, y(t) = e^{3t –}– e^2t

given, l[f(t)] = s + 2/s² + 1 = f(s)

l[g(t)] = s² + 1/(s + 3)(s + 2) = g(s)

h(t) = f g(t-) d

h(t) = f(t)* g(t)

taking laplace transform,

l[h(t) = l[f(t)* g(t)]

= l[f(t)]. l[g(t)]

= f(s). g(s)

= s + 2/s² + 1 x s² + 1/(s + 3) (s + 2) = 1/s + 3

the signal x(t) = e^-3t2 is an example of gaussian distribution. the fourier transform of gaussian distribution is also caussian distribution.

x(f) = A e^-bf2

let x(t) = a x₁ (t) + b x₂ (t)

then, a y₁ (t) + b y₂ (t) = at x₁ (t) + bt x₂(t)

= t x(t)

then system is linear.

if input is delayed by t₀, i.e., x(t – t₀)

then output is

y₁ (t) = t x (t – t_o)

but output is delayed by t_o.

y₂(t) = (t – t_o x (t – t_o)

y₁ (t) = y₂(t)

hence, system is not time-invariant, i.e., time varying.

x[n] = x_dft [k]

a typial DFT term X_DFT[K] has the form Ae^jo

let = 2kn /n, the IDFT gives x[n] with terms of the form

1/n Ae^jo . e^j = 1/n Ae^{j(0+ 0)}

if we conjugate X_DFT [k], typical term becomes Ae^-jo. its IDFT gives terms of the form Ae^-joe^-jo = Ae^{-j(0 + 0)}

this DFT result corresponds to the signal Nx *[n].

the DFT of a real signal shows conjugate symmetry as

X_DFT[K] = X_DFT [N – K]

here, N = 6

a = X_DFT [1] = X_DFT [6 – 1] = X_DFT [5]

a = – j

B = X_DFT [4] = X_DFT[6 – 4] = X_DFT[2]

B = 2 – J

let x[n] = (1,2,1,2} is discrete fourier transform of x[n] is given as

X_DFT [K] = X[n] e^-j2nk/n

k = 0, 1,……….N – 1 and

n = 4

so, X_DFT [K] = x[n] e^-j2nk/2

x[n] e^-jkn/2

k = 0, X_DFT [0] = x[n] e^o

x[0] + x[1] + x[2] + x[3]

= 1 + 2 + 1 + 2 = 6

k = 1, X_DFT [1] = x[n] e^-jn/2

= x₁ [0] e⁰ + x[1] e^-j/2 + x[2] e^-j + x[3] e^j3/2

= 1 – 2j – 1 + 2j = 0

k = 2, X_DFT [2] = x[n] e^-jn

= x[0] e⁰ + x[1] e^-j + x[2] e^-j2 + x[3] e^-j3

= 1 – 2 + 1 – 2 = 2

k = 3, X_DFT [3] = x[n] e^-jn3/2

= x[10] e⁰ + x [1] e^-j3/2 + x[2] e^-j3 + x[3] e^-j9/2

= 1 – 2j – 1 + 2j = 0

so, X_DFT [K] = {6,0, -2, 0}

for a band-pass signal, minimum sampling frequency is given as

f_s = 2fH/N

where, N = int (fH/B)

Here, f_l = 4 kHz, f_h = 6 kHz

bandwidth B = F_H – F_L = 6 – 4 = 2 kHz

N = int (6/2) = 3

f_s = 2 x 6/3 = 4 kHz

x₁ (t) * x₂(t) = x₁ (f) x₂ (f) (product of spectrum)

so, x₁(t)* x₂(t) extends only to 2 kHz

nyquist rate = 2 x 2 kHz = 4 kHz

Because present output y(n) depends upon past output y(n – 1) and future output y(n + 1).

for causal system output must depend upon present and past, not on future.

The periods of individual components are

0₁ = 2/3 2/T₁ = 2/3 t₁ = 3

0₂ = 1/4 2/t₂ = 1/4 t₂ = 8

0₃ = 1/3 2/t₃ = 1/3 t₂ = 6

0₄ = 1/2 2/t₄ = 1/2 t₄ = 4

Common period is LCM of the individual periods

T = LCM (3, 8, 6, 4)

= 24